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Problem of the Week
Problem D and Solution
Halfway to the Other Side


Cube \(PQRSTUVW\) has side length \(2\). Point \(M\) is the midpoint of edge \(UT\). Determine the area of \(\triangle MQR\).

The top face of the cube has vertices P, Q, R, S. On the
bottom face, U is below P, V is below Q, W is below R, and T is below S.
UT, with midpoint M, and QR are diagonally opposite edges. Sides MQ and
MR of triangle MQR pass through the interior of the cube.


We first draw \(RT\).

RT is the diagonal of cube face RSTW.

In \(\triangle RWT\), \(\angle RWT = 90^\circ\) and \(RW = WT = 2\).

By the Pythagorean Theorem in \(\triangle RWT\), \(RT^2= RW^2 + WT^2= 2^2 + 2^2=8\).

Therefore, \(RT = \sqrt{8}\), since \(RT>0\).

\(\triangle MQR\) has base equal to the length of \(QR\), which is \(2\).

Notice that the height of \(\triangle MQR\) is equal to the distance from side \(QR\) of the cube to side \(UT\) of the cube, which is equal to the length of \(RT\) or \(\sqrt{8}\).

Therefore, area of \(\triangle MQR = \dfrac{\text{base}\times \text{height}}{2} = \dfrac{2 \times \sqrt{8}}{2}= \sqrt{8}\) units squared.